Arock falls from rest a vertical distance of 0.72 meters to the surface of a planet in 0.63 seconds. the magnitude of the acceleration due to gravity on the planet ?

3.63 meter/sec²

Step-by-step explanation:

When a rock fall from a height h with initial velocity u and the rock reaches the surface in t seconds the expression that represents this is

h = 4t + gt²

Where g is the acceleration due to gravity

Here h = 0.72 meters

        u = 0

        t = 0.63 seconds

from the given formula

0.72 = 0 ×(0.63) + g(0.63)²

0.72 = g(0.63)²

g =

= 3.63 meter/sec²

3.63 meter/sec²

Step-by-step explanation:

If a rock fall from a height h, then

Where g is the acceleration due to gravity, u is initial velocity and t is time in seconds in which the rock reaches the surface.

It is given that a rock falls from rest a vertical distance of 0.72 meters to the surface of a planet in 0.63 seconds.

h = 0.72 meters

u = 0

t = 0.63 seconds

Substitute the given values in the above formula to find the value of g.

Multiply both sides by 2.

Divide both sides by 0.3969.

Therefore, the magnitude of the acceleration due to gravity on the planet is  3.63 meter/sec².

v: l•w•h÷2.

v: 20•6•8 ÷2

v: 960÷2

v: 480